What is the solution for x^{''}+9x=10cos(2t) ?

The solution for x^{''}+9x=10cos(2t) is x=c_{1}cos(3t)+c_{2}sin(3t)+2cos(2t)

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x^{''}+9x=10cos(2t)

x^{^{''}} + 9 x = 10 cos (2 t)

x = c_{1} cos (3 t) + c_{2} sin (3 t) + 2 cos (2 t)

Solution steps

x^{''} (t) + 9 x = 10 cos (2 t)

Solve linear ODE:

x = c_{1} cos (3 t) + c_{2} sin (3 t) + 2 cos (2 t)

x = c_{1} cos (3 t) + c_{2} sin (3 t) + 2 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 10 y = 30 t + 9, y (0) = 7, y^{^{'}} (0) = 26

- 4 y^{^{''}} + y^{^{'}} + 9 y = 15

y^{^{''}} + 50 y^{^{'}} + 100 = 0

y^{^{''}} - 3 y^{^{'}} = - 18 x, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 8 y^{^{'}} + 17 y = x^{2} e^{4 x}

What is the solution for x^{''}+9x=10cos(2t) ?
The solution for x^{''}+9x=10cos(2t) is x=c_{1}cos(3t)+c_{2}sin(3t)+2cos(2t)